CHAPTER 53 METHODS OF DIFFERENTIATION
EXERCISE 225 Page 617
1. Differentiate with respect to x: (a) 5x5
(a) If y = 5x5 then
1
x
dy
= (2.4) ( 3.5 x3.5−1 ) = 8.4x 2.5
dx
dy
1
1
= −1 x −2 = −
= x −1 then
x2
dx
x
2. Differentiate with respect to x: (a)
(a) If y =
(c)
dy
= (5) ( 5 x 4 ) = 25x 4
dx
(b) If y = 2.4x3.5 then
(c) If y =
(b) 2.4x3.5
−4
x2
(b) 6
(c) 2x
−4
dy
8
= −4x −2 then
( 4) ( −2 x −3 ) =
8 x −3 or
=−
2
x
dx
x3
(b) If y = 6 then
(c) If y = 2x then
dy
=0
dx
dy
=2
dx
3. Differentiate with respect to x: (a) 2 x
(b) 3 3 x5
(c)
4
x
1
1
1
dy
1
 1 − 12 
−
2
(a) If y = 2 x = 2 x 2 then = (2)  x =
=
 x=
1
dx
x
2

x2
2
5
dy
 5 23 
3 = 5 3 x2
(b) If y = 3 3 x5 = 3 x 3 then
= (3)
=
x
5
x


dx
3


1
3
2
4
4
dy
2
−
 1 − 32 
−
2 then
2 =
(c) If y = = =
4x
=−
(4)
x
=
−
2
x
− 3 =−


1
dx
x3
x x2
 2

x2
911
© 2014, John Bird
4. Differentiate with respect to x: (a)
(a) If y = −
−3
x
3
(b) (x – 1)2
(c) 2 sin 3x
1
4
3
3
dy
1
−
−
 1 − 43 
3 then
3 =
=
=
−
=
−
3x
=
−
−
=
x
x
(
3)


1
4
3
x
d
3
x


x3
x3
3
1
x4
(b) If y = (x – 1)2 = (x – 1)(x – 1) = x 2 − x − x + 1 = x 2 − 2 x + 1
then
dy
= 2x – 2 or 2(x – 1)
dx
(c) If y = 2 sin 3x
dy
= (2)(3 cos 3x) = 6 cos 3x
dx
then
(b) 2e6x
5. Differentiate with respect to x: (a) – 4 cos 2x
then
(b) If y = 2e6x then
dy
= (2) ( 6 e6 x ) = 12 e6 x
dx
dy
3
15
=
(3) ( −5e −5 x ) =
−15e −5 x = −
= 3e −5 x then
e5 x
dx
e5 x
6. Differentiate with respect to x: (a) 4 ln 9x
(a) If y = 4 ln 9x
then
(b)
e x − e− x
2
(c)
1− x
x
dy
1 4
= (4)   =
x
dx
x
(b) If y =
e x − e− x 1 x 1 − x
=
e − e
2
2
2
(c) If y =
1
1
1− x 1
x
−1
−
= −
= x −1 − x 2 = x −1 − x 2
x
x x
then
3
e5 x
dy
= (–4)(– 2 sin 2x) = 8 sin 2x
dx
(a) If y = –4 cos 2x
(c) If y =
(c)
then
dy 1 x 1
1
1
e x + e− x
=   e −   ( − e x=
)   e x +   e− x =
dx 2
2
2
2
2
dy
1 1 −3
1
1
1
1
 1 − 1 −1 
=
− x −2 −  − x 2  =
− + x 2=
− + 3 =− +
dx
x2 2
x2 2 x 2
x 2 2 x3
 2

912
© 2014, John Bird
7. Find the gradient of the curve y = 2t4 + 3t3 – t + 4 at the points (0, 4) and (1, 8)
If y= 2t 4 + 3t 3 − t + 4 , then gradient,
dy
= 8t 3 + 9t 2 − 1
dt
At (0, 4), t = 0, hence gradient = 8(0)3 + 9(0) 2 − 1 = – 1
At (1, 8), t = 1, hence gradient = 8(1)3 + 9(1) 2 − 1 = 16
8. Find the coordinates of the point on the graph y = 5x2 – 3x + 1 where the gradient is 2
dy
If y = 5 x 2 − 3 x + 1 , then gradient = = 10 x − 3
dx
When the gradient is 2, 10x – 3 = 2
i.e. 10x = 5
and x =
1
2
2
5 3
3
1
1
y = 5  − 3  +1 = − +1 =
4 2
4
2
2
1
When x = ,
2
1 3
Hence, the coordinates of the point where the gradient is 2 is  , 
2 4
9. (a) Differentiate y =
(b) Evaluate
(a)
2
θ
2
+ 2 ln 2θ – 2(cos 5θ + 3 sin 2θ) –
2
e3θ
dy
π
when θ = , correct to 4 significant figures.
dθ
2
2
2
y = + 2 ln 2θ − 2(cos 5θ + 3sin 2θ ) −
θ2
e3θ
= 2θ −2 + 2 ln 2θ − 2 cos 5θ − 6sin 2θ − 2e −3θ
Hence,
dy
2
=
−4θ −3 + − 2(−5sin 5θ ) − 6(2 cos 2θ ) − 2 ( −3e −3θ )
dθ
θ
=−
(b) When θ =
π
2
,
4
θ3
+
2
θ
+ 10sin 5θ − 12 cos 2θ +
6
e3θ
dy
4
2
5π
2π
6
=−
+
+ 10sin
− 12 cos
+ π 
3
3 
dθ
2
2
π  π 
e 2
   2 
2
= –1.032049 + 1.2732395 + 10 + 12 + 0.0538997
913
© 2014, John Bird
= 22.30, correct to 4 significant figures
10. Evaluate
ds
π
, correct to 3 significant figures, when t =
given s = 3 sin t – 3 +
dt
6
t.
1
ds
1 −1
1
2
3cos t +
If s = 3 sin t – 3 + =
t 3sin t − 3 + t 2 then = 3cos t − 0 + t=
dt
2
2 t
π ds
π 
When t ==
,
3cos   +
6 dt
6 2
1
(π / 6 )
= 3.29, correct to 3 significant figures
11. A mass m is held by a spring with a stiffness constant k. The potential energy p of the system
is given by:=
p
1 2
kx − mgx where x is the displacement and g is acceleration due to gravity.
2
The system is in equilibrium if
If p
=
If
dp
= 0 . Determine the expression for x for system equilibrium.
dx
1 2
dp
kx − mgx then
= kx − mg
2
dx
dp
= 0 then kx – mg = 0 i.e.
dx
kx = mg
and displacement, x =
mg
k
12. The current i flowing in an inductor of inductance 100 mH is given by: i = 5 sin 100t amperes,
where t is the time t in seconds. The voltage v across the inductor is given by: v = L
di
volts.
dt
Determine the voltage when t = 10 ms.
di
If i = 5 sin 100t=
then
(5)(100
=
cos t ) 500 cos100t
dt
Voltage across inductor, v = L
di
= ( L)(500 cos100t )
dt
When L = 100 mH and t = 10 ms, voltage, v = (100 ×10−3 )( 500 cos(100 ×10 ×103 ) )
= 50 cos 1 = 27.0 volts
914
© 2014, John Bird
EXERCISE 226 Page 618
1. Differentiate with respect to x: x sin x
dy
If y = x sin x, then
=
dx
( x )( cos x ) + ( sin x )(1)
= x cos x + sin x
2. Differentiate with respect to x: x 2 e 2 x
dy
If y = x 2 e 2 x ,=
then
dx
( x 2 )( 2 e2 x ) + ( e2 x )( 2 x )
= 2 x 2 e 2 x + 2 x e 2 x or 2 x e 2 x ( x + 1)
3. Differentiate with respect to x: x 2 ln x
dy
=
If y = x 2 ln x , then
dx
( x 2 ) 
1
 + ( ln x )( 2 x )
x
= x + 2x ln x or
x(1 + 2 ln x)
4. Differentiate with respect to x: 2x3 cos 3x
If y = 2 x 3 cos 3 x , then
dy
=( 2 x3 )( −3sin 3 x ) + ( cos 3 x )( 6 x 2 )
dx
= −6 x3 sin 3 x + 6 x 2 cos 3 x = 6 x 2 ( cos 3 x − x sin 3 x )
5. Differentiate with respect to x:
If y =
x3 ln 3x
3
d y  3  1 
3 1
=  x 2    + ( ln 3 x )  x 2 
x3 ln 3x = x 2 ln 3 x , then
d x   x 
2 
3
−1
= x2 +
1
3 1
3 1
x 2 ln 3x =
x 2 + x 2 ln 3x
2
2
1
 3

= x 2 1 + ln 3 x  =
 2

915
 3

x 1 + ln 3 x 
 2

© 2014, John Bird
6. Differentiate with respect to x: e3t sin 4t
dy
If y = e3t sin 4=
t , then
dt
( e3t )( 4 cos 4t ) + ( sin 4t )( 3e3t )
= e3t ( 4 cos 4t + 3sin 4t )
7. Differentiate with respect to θ: e4θ ln 3θ
dy
If y = e4θ ln 3θ, then
=
dθ
( e4θ ) 
1
4θ
 + ( ln 3θ )( 4 e )
θ 
1

= e 4θ  + 4 ln 3θ 
θ

8. Differentiate with respect to t: et ln t cos t
If y = et ln t cos t , then
dy
=
dt


1
t
 + ( ln t )( e ) 
t 

( et ln t )( − sin t ) + ( cos t ) ( et ) 

cos t


+ cos t ln t 
= et − ln t sin t +
t


 1


= et  + ln t  cos t − ln t sin t 

 t

9. Evaluate
di
, correct to 4 significant figures, when t = 0.1, and i = 15 t sin 3t
dt
di
Since i = 15t sin=
3t, then
(15t )(3cos 3t ) + (sin 3t )(15)
dt
= 45t cos 3t + 15 sin 3t
When t = 0.1,
di
= 45(0.1) cos 0.3 + 15 sin 0.3
dt
(note 0.3 is radians)
= 4.2990 + 4.4328
= 8.732, correct to 4 significant figures
10. Evaluate
dz
, correct to 4 significant figures, when t = 0.5, given that z = 2e3t sin 2t
dt
916
© 2014, John Bird
dz
Since z = 2e3t sin
2t, then
=
(2 e3t )(2 cos 2t ) + (sin 2t )(6 e3t )
dt
= 4 e3t cos 2t + 6 e3t sin 2t
When t = 0.5,
dz
= 4 e1.5 cos1 + 6 e1.5 sin1
dt
(note 1 is 1 radian)
= 9.68587 + 22.62727
= 32.31, correct to 4 significant figures
917
© 2014, John Bird
EXERCISE 227 Page 620
sin x
x
1. Differentiate with respect to x:
If y =
sin x
, then
x
d y ( x )( cos x ) − ( sin x )(1)
=
2
dx
( x)
=
x cos x − sin x
x2
2 cos 3x
x3
2. Differentiate with respect to x:
x 3 )( −6sin 3x ) − ( 2 cos 3 x )( 3 x 2 )
(=
2
( x3 )
dy
2 cos 3x
If y =
, then
=
dx
x3
=
−6( x sin 3 x + cos 3 x)
x4
dy
2x
,=
then
2
dx
x +1
x6
−6
( x sin 3x + cos 3x )
x4
2x
x2 + 1
3. Differentiate with respect to x:
If y =
or
−6 x 2 ( x sin 3x + cos 3x )
( x 2 + 1)( 2=
) − ( 2 x )( 2 x )
2
( x 2 + 1)
2 x2 + 2 − 4 x2
=
2
( x 2 + 1)
2 − 2 x2
( x 2 + 1)
=
x
x
If y =
, then
=
cos x cos x
2(1 − x 2 )
( x 2 + 1)
2
x
cos x
4. Differentiate with respect to x:
1
2
2
dy
=
dx
( cos x ) 
1 −1 
x 2 −
2

( x ) ( − sin x )
( cos x )
2
cos x
+ x sin x
2
x
=
cos 2 x
918
© 2014, John Bird
3 θ3
2sin 2θ
5. Differentiate with respect to θ:
If y =
3
2
dy
3 θ3
3θ
, then
=
=
dθ
2sin 2θ 2sin 2θ
( 2sin 2θ ) 
9 1  3
θ 2  −  3θ 2  ( 4 cos 2θ )

2  
( 2sin 2θ )
2
3 θ { 3sin 2θ − 4θ cos 2θ }
9 θ 2 sin 2θ − 12 θ 2 cos 2θ
=
=
4sin 2 2θ
4sin 2 2θ
1
6. Differentiate with respect to t:
3
ln 2t
t
( t )  1t  − ( ln 2t )  12 t 
=
( t)
−
If y =
ln 2t
dy
,=
dt
t
1
2
7. Differentiate with respect to x:
2 x e4 x
, then
sin x
1
2
1 −1
1
−
− t 2 ln 2t
2 = t 2
t
t
 1

1 − ln 2t 
 2

1  1
1 1


1 − ln 2t 
1 − ln 2t  =

3 
t3  2


t2  2
2 xe 4 x
sin x
d y ( sin x ) ( 2 x )( 4 e 4 x ) + ( e 4 x )( 2 )  − ( 2 x e 4 x )( cos x )
=
2
dx
( sin x )
=
8 x e 4 x sin x + 2 e 4 x sin x − 2 x e 4 x cos x
sin 2 x
=
2 e4 x
{(1 + 4 x ) sin x − x cos x}
sin 2 x
8. Find the gradient of the curve y =
If y =
−
2
=
If y =
t
dy
2x
then =
gradient,
2
dx
x −5
2x
at the point (2, –4)
x −5
2
− (2 x)(2 x)
( x 2 − 5) (2)
=
2
( x 2 − 5)
At the point (2, –4), x = 2, hence gradient =
2 x 2 − 10 − 4 x 2 −10 − 2 x 2
=
2
2
( x 2 − 5)
( x 2 − 5)
−10 − 2(2) 2 −10 − 8 −18
= –18
=
=
2
(4 − 5) 2
1
( 22 − 5 )
919
© 2014, John Bird
9. Evaluate
If y =
dy
2x2 + 3
at x = 2.5, correct to 3 significant figures, given y =
ln 2 x
dx
dy
2x + 3
then
=
dx
ln 2 x
2
dy
When x = 2.5,
=
dx
( ln 2 x ) (4 x) − (2 x 2 + 3) 
( ln 2 x )
1

x
2
( ln 5) (10) − [2(2.5)2 + 3] 
1 

 2.5  16.09438 − 6.2 = 3.82, correct to 3
=
2
2.59029
( ln 5)
significant figures
920
© 2014, John Bird
EXERCISE 228 Page 621
( 2 x − 1)
1. Differentiate with respect to x:
6
dy
6
5
If y = ( 2 x − 1) then = 6 ( 2 x − 1) ( 2 )
dx
dy
5
i.e. = 12 ( 2 x − 1)
dx
2. Differentiate with respect to x: (2x3 – 5x)5
If y = ( 2 x3 − 5 x ) then
5
dy
4
=
5 ( 2 x3 − 5 x ) ( 6 x 2 − 5)
dx
3. Differentiate with respect to θ: 2 sin(3θ – 2)
dy
If y = 2 sin(3θ – 2) then
= 2 ( cos ( 3θ − 2 ) ) ( 3)
dθ
dy
i.e.
= 6 cos ( 3θ − 2 )
dθ
4. Differentiate with respect to α: 2 cos5 α
dy
If y = 2 cos5 α=
then
(2) ( 5cos 4 α )( − sin α )
dα
1
( x3 − 2 x + 1)
5
= ( x3 − 2 x + 1)
dy
= −10 cos 4 α sin α
dα
1
( x − 2 x + 1)5
5. Differentiate with respect to x:
If y =
i.e.
3
−5
then
−5 ( 3 x 2 − 2 )
dy
−6
=
−5 ( x3 − 2 x + 1) ( 3 x 2 − 2 ) =
6
dx
( x3 − 2 x + 1)
=
5 ( 2 − 3x 2 )
( x3 − 2 x + 1)
6
6. Differentiate with respect to t: 5e2t+1
If y = 5e2t+1 then
dy
= (5) ( e 2t +1 )( 2 )
dt
i.e.
dy
= 10 e 2t +1
dt
921
© 2014, John Bird
7. Differentiate with respect to t: 2 cot(5t2 + 3)
cos x
then
sin x
If y = cot x =
d y (sin x)(− sin x) − (cos x)(cos x) − ( sin 2 x + cos 2 x )
−1
=
=
=
= − cosec 2 x
dx
sin 2 x
sin 2 x
sin 2 x
Thus, if y = 2 cot ( 5t 2 + 3) , then
dy
=
( 2 )  − cos ec 2 ( 5t 2 + 3) (10t ) = −20t cosec2 ( 5t 2 + 3)
dt
8. Differentiate with respect to y: 6 tan(3y + 1)
dθ
(6) ( sec 2 ( 3 y + 1) ) ( 3)
If θ = 6 tan(3y + 1) =
then
dy
dθ
=
18sec 2 ( 3 y + 1)
i.e.
dy
9. Differentiate with respect to x: 2etan θ
If y = 2 e tan θ , then
dy
= ( 2e tan θ ) sec 2 θ = 2sec 2 θ e tan θ
dθ
10. Differentiate: θ sin(θ –
θ=
π
π
3
) with respect to θ, and evaluate, correct to 3 significant figures, when
2
π
dy

If y = θ sin  θ −  then =
dθ
3

When θ =
π
2
,
(θ ) cos  θ −

π
π

 + sin  θ −  (1)
3
3

d y π 
π
π
π π 
π π  π
=   cos  −  + sin  − =
cos + sin

dθ  2 
6
6
2 3
2 3 2
= 1.360 + 0.5 = 1.86, correct to 3
significant figures
922
© 2014, John Bird
11. The extension, x metres, of an undamped vibrating spring after t seconds is given by:
x = 0.54 cos(0.3t – 0.15) + 3.2
Calculate the speed of the spring, given by
If x = 0.54 cos(0.3t – 0.15) + 3.2 then
dx
, when (a) t = 0, (b) t = 2 s
dt
dx
= (0.54) [– sin(0.3t – 0.15)](0.3) + 0
dt
= –0.162 sin(0.3t – 0.15)
(a) When t = 0, speed of spring,
(b) When t = 2 s, speed of spring,
dx
= –0.162 sin(–0.15) = 0.02421 m/s = 24.21 mm/s
dt
dx
= –0.162 sin(0.45) = –0.07046 m/s = –70.46 mm/s
dt
923
© 2014, John Bird
EXERCISE 229 Page 622
1. If y = 3x4 + 2x3 – 3x + 2 find (a)
d2 y
d x2
(a) If y = 3x4 + 2x3 – 3x + 2 then
dy
d2 y
= 12 x3 + 6 x 2 − 3 and = 36 x 2 + 12 x
d x2
dx
(b)
(b)
d3 y
d x3
d3 y
= 72 x + 12
d x3
2. (a) Given f(t) =
2 2 1
3
t –
+ –
5
t3
t
t + 1 determine f ´´(t)
(b) Evaluate f ´´(t) when t = 1
(a) f(t) =
1
2 2 1 3
2 2 −3
t − + − t +=
1
t − t + 3t −1 − t 2 + 1
5
t3 t
5
f ′(t) =
4
1 −1
t + 3t −4 − 3t −2 − t 2
5
2
f ′′(t) =
4 12 6
1
4
1 −3
− 12t −5 + 6t −3 + t 2 = − + +
5
3
5 t
t
5
4
4 t3
(b) When t = 1, f ′′(t) =
4 12
6
1
4
1
−
+
+
= − 12 + 6 + = –4.95
5
3
3
5 (1) (1) 4 1
5
4
3. The charge q on the plates of a capacitor is given by q = CV e
−
t
CR
, where t is the time, C is the
capacitance and R the resistance. Determine (a) the rate of change of charge, which is given by
dq
d2 q
, (b) the rate of change of current, which is given by
dt
d t2
(a) If q = CV e
(b) If
−
t
CR
dq
V − t
 1 − CRt 
e
,= (CV )  −
 = − e CR
dt
R
 CR

dq
V − t
V − t d 2 q  V  1 − t 
=
−  −
e CR  =
e CR
= − e CR ,

d t 2  R  CR
dt
R
 CR 2
924
© 2014, John Bird
4. Find the second differential coefficient with respect to the variable:
(a) 3 sin 2t + cos t
(a) If y = 3 sin 2t + cos t
and
then
dy
= (3)(2 cos 2t ) − sin=
t 6 cos 2t − sin t
dx
d2 y
= (6)(– 2 sin 2t) – cos t = – 12 sin 2t – cos t or –(12 sin 2t + cos t)
d x2
(b) If y = 2 ln 4θ then
and
(b) 2 ln 4θ
dy
1 2
−1
= (2)  =
 = 2θ
dx
θ  θ
d2 y
2
= −2θ −2 = −
2
dx
θ2
5. Find the second differential coefficient with respect to the variable:
(a) 2 cos2x
(b) (2x – 3)4
(a) If y = 2 cos 2 x ,
dy
=
4 cos x(− sin x) =
−4sin x cos x
dx
d2 y
=
(−4sin x)(− sin x) + (cos x)(−4 cos x) =
4sin 2 x − 4 cos 2 x
d x2
and
= 4 ( sin 2 x − cos 2 x )
(b) If y = ( 2 x − 3) ,
4
and
dy
=4(2 x − 3)3 (2) =8(2 x − 3)3
dx
d2 y
2
= 24(2 x − 3) 2 (2) = 48 ( 2 x − 3)
d x2
6. Evaluate f ´´(θ) when θ = 0 given f(θ) = 2 sec 3θ
If f(θ) = 2 sec 3θ, then f ′(θ) = 6 sec 3θ tan 3θ
and
f ′′(θ) = ( 6sec 3θ )( 3sec 2 3θ ) + ( tan 3θ )(18sec 3θ tan 3θ )
= 18sec3 3θ + 18sec 3θ tan 2 3θ
When θ = 0,
f ′′(0) =
18
18 tan 2 0 18 18(0)
= 18
+
=
+
cos3 0
cos 0
1
1
925
© 2014, John Bird
7. Show that the differential equation
If y = xe2x then
d2 y
dy
–4
+ 4y = 0 is satisfied when y = xe2x
2
dx
dx
dy
=( x)(2 e 2 x ) + (e 2 x )(1) =2 x e 2 x + e 2 x
dx
d2 y
= (2 x)(2 e 2 x ) + (e 2 x )(2) + 2 e 2 x = 4 x e 2 x + 4 e 2 x
d x2
and
dy
d2 y
–4
+ 4y = ( 4 xe 2 x + 4e 2 x ) – 4( 2 x e 2 x + e 2 x ) + 4(xe2x)
2
dx
dx
= 4 xe 2 x + 4e 2 x – 8 x e 2 x − 4 e 2 x + 4xe2x
=0
8. Show that, if P and Q are constants and y = P cos(ln t) + Q sin(ln t), then t 2
d2 y
dy
+t
+y=0
2
dt
dt
y = P cos(ln t) + Q sin(ln t)
− P sin(ln t ) + Q cos(ln t )
P
Q
dy
=
− sin(ln t ) + cos(ln t ) =
t
t
t
dt
 − P cos(ln t ) Q sin(ln t ) 
−
(t ) 
 − [ − P sin(ln t ) + Q cos(ln t ) ] (1)
d y
t
t

=
t2
d t2
2
=
− P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t )
t2
Hence, t 2
d2 y
dy
− P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t )
+t
+ y = (t 2 )
2
dt
dt
t2
+ (t)
− P sin(ln t ) + Q cos(ln t )
+ P cos(ln t) + Q sin(ln t)
t
= − P cos(ln t ) − Q sin(ln t ) + P sin(ln t ) − Q cos(ln t ) − P sin(ln t ) + Q cos(ln t )
+ P cos(ln t) + Q sin(ln t)
=0
926
© 2014, John Bird
9. The displacement s of a mass in a vibrating system is given by: s=
natural frequency of vibration. Show that:
If
s=
and
(1 + t ) e−ω t
(1 + t ) e−ω t
where ω is the
d2 s
ds
0
+ 2ω
+ ω 2s =
2
dt
dt
ds
=+
(1 t ) ( −ω e −ωt ) + (e −ωt )(1) =
−ω e −ωt − ωt e −ωt + e −ωt
dt
d2 y
=
+ω 2 e − ωt − [(ωt ) ( −ω e − ωt ) + ( e −ωt ) (ω )] − ω e −ωt
d x2
= ω 2 e − ωt + ( ω 2 t e − ωt ) − ( ω e − ωt ) − ω e − ωt
= ω 2 e − ωt + (ω 2t e − ωt ) − 2ω e −ωt
Hence,
d2 s
ds
ω 2 e − ωt + (ω 2t e − ωt ) − 2ω e −ωt + 2ω[−ω e −ωt − ωt e −ωt + e −ωt ] + ω 2 [(1 + t ) e −ωt ]
+ 2ω
+ ω 2s =
2
dt
dt
= ω 2 e − ωt + ω 2te −ωt − 2ω e −ωt − 2ω 2 e −ωt − 2ω 2te −ωt + 2ω e −ωt + ω 2 e −ωt + ω 2te −ωt
=0
927
© 2014, John Bird